Is the following definition of continuity accurate?

Question

Is the following (rather wordy) definition of continuity accurate?

Let $f$ be a function with domain $A$, $ $ then $\forall x \in A$ $ $ and $ $ for $c \in A$, $ $ $f$ is continuous at $c$;

if when $\delta>0$ is the length of the interval $[x, c]$, for $x<c$, $ $ or the length of the interval $[c, x]$, for $c<x$,
there exists $\varepsilon>0$, $ $ such that $\varepsilon$ is the length of the interval $[f(x), f(c)]$, $ $ if $f(x)<f(c)$, $ $ or the length of the interval $[f(c), f(x)]$, $ $ if $f(c)<f(x)$.

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EDITED version (after the comments and answers below):

Let $f$ be a function with domain $A$, $ $ then $\forall x \in A$ $ $ and $ $ for $c \in A$, $ $ $f$ is continuous at $c$;

if when $\delta>0$ is the length of the interval $[x, c]$, for $x<c$, $ $ or the length of the interval $[c, x]$, for $c<x$,
there exists $\varepsilon>0$, $ $ such that $\varepsilon$ is the length of the interval $[f(x), f(c)]$, $ $ if $f(x)<f(c)$, $ $ or the length of the interval $[f(c), f(x)]$, $ $ if $f(c)<f(x)$, $ $ and this interval is a subset of $f(A)$.

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Just trying to rephrase this in my own words to make sure I understand the actual definition.

The key idea here is the intervals, so I could get rid of the $\varepsilon$ and the $\delta$, to simplify it a bit, but that would defeat the purpose of the "$\varepsilon$ - $\delta$" definition, I suppose.

Answer 1

Not quite. See, the mathematical definition has an $\epsilon$ before a $\delta$, and your statement goes : "If, when $\delta$ is ... , there exists $\epsilon > 0$ ...".

The mathematical definition goes roughly like : "If, when $\epsilon > 0$ is ..., there exists $\delta > 0$ ...", so you have got the two mixed up, a highly common mistake.

If I were to write the definition of continuity for you, then I would do it as follows:

Given a function $f$, with domain $A$ and a point $c \in A$, $f$ is continuous at the point $c$, if: for every $\epsilon >0$, there is a $\delta > 0$, such that for every point $x$, if the distance between $x$ and $c$ is smaller than $\delta$, then the distance between $f(x)$ and $f(c)$ is smaller than $\epsilon$.

Note the order of variables in this definition(forget $f,A,c$ : these are part of the definition, but they match up in the definition you and I provide and are correctly provided, so you may ignore these).

• First, comes $\epsilon > 0$. Usually, this step is phrased as "given $\epsilon > 0$". The word given is important, since here we don't get to decide $\epsilon$, but rather somebody(not influenced by us) gives us an $\epsilon > 0$ and (without protest) we must work with it.

• Then, comes the $\delta > 0$, which possibly(note : in some cases it may not) depends on $\epsilon$, and is your decision to make. That is, you may have to decide your $\delta$ based on the $\epsilon > 0$ that is given to you.

• How must you decide the $\delta$? This is given as follows : for every $x$ which is in the domain, and whose distance from $c$ is less than $\delta$, we must have that the distance between $f(x)$ and $f(c)$ is less than $\epsilon$.

That is, if some point is $\delta$-close to $c$, then its image under $f$ is $\epsilon$-close to $f(c)$.

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So let's correct your statement. The $\epsilon - \delta$ order must be restored, and the intervals must be made two-sided, since we are only saying above that $x$ is close to $c$, not to the left or right. An interval of the form $[c,x]$ is indicative of such an order, and is therefore a problem in the definition, one that must be avoided in any rephrasing.

Given a function $f$ with domain $A$ and a point $c \in A$, $f$ is said to be continuous at $c$, if given $\epsilon > 0$, we can find a $\delta > 0$, such that for all $x$ in the domain satisfying $|x-c| < \delta$, $x$ will also satisfy $|f(x) - f(c)| < \epsilon$.

Once again, I request you to observe the order of the variables in this definition. This is all you have got wrong,so good effort nevertheless.

Answer 2

That's wrong. Take$$\begin{array}{rccc}\chi_{\mathbb Q}\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x\in\mathbb Q\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $\chi_{\mathbb Q}$ is discontinuous everywhere, but it would be continuous according to your definition (for each $\delta>0$, take $\varepsilon=2$).

Answer 3

This is nice interpretation but wrong.

$$f(x)=\begin{cases}x\text{ everywhere apart from $0$}\\\text{undefined at $0$}\end{cases}$$

Let's choose $c=1$ and $x=-1$, so we have the interval $[-1,1]$ but the interval $[f(-1),f(1)]$ have a point outside of the image, although the function is continuous at $1$.

The problem is that the initial part is $\delta$ and not $\varepsilon$.

Now consider the following: for any $\varepsilon>0$ exists some $\delta>0$ such that if $x$ is in domain and in $[c-\delta,c+\delta]$ then $f(x)$ is in the interval $[f(c)-\varepsilon,f(c)+\varepsilon]$.

This is different because you can see that $\delta$ depends on $\varepsilon$, maybe it is clearer like this.

In words it means: we say that $f(x)$ is continuous at $c$ if for any distance from $f(c)$ we take we there exists a distance from $c$ such that if we are within the distance from $c$ we are also within the distance from $f(c)$